3.20.15 \(\int \frac {x^4}{2 x+13 x^2+15 x^3} \, dx\)

Optimal. Leaf size=33 \[ \frac {x^2}{30}-\frac {13 x}{225}+\frac {8}{189} \log (3 x+2)-\frac {1}{875} \log (5 x+1) \]

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1585, 701, 632, 31} \begin {gather*} \frac {x^2}{30}-\frac {13 x}{225}+\frac {8}{189} \log (3 x+2)-\frac {1}{875} \log (5 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(2*x + 13*x^2 + 15*x^3),x]

[Out]

(-13*x)/225 + x^2/30 + (8*Log[2 + 3*x])/189 - Log[1 + 5*x]/875

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2,
0] && NeQ[2*c*d - b*e, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rubi steps

\begin {align*} \int \frac {x^4}{2 x+13 x^2+15 x^3} \, dx &=\int \frac {x^3}{2+13 x+15 x^2} \, dx\\ &=\int \left (-\frac {13}{225}+\frac {x}{15}+\frac {26+139 x}{225 \left (2+13 x+15 x^2\right )}\right ) \, dx\\ &=-\frac {13 x}{225}+\frac {x^2}{30}+\frac {1}{225} \int \frac {26+139 x}{2+13 x+15 x^2} \, dx\\ &=-\frac {13 x}{225}+\frac {x^2}{30}-\frac {3}{175} \int \frac {1}{3+15 x} \, dx+\frac {40}{63} \int \frac {1}{10+15 x} \, dx\\ &=-\frac {13 x}{225}+\frac {x^2}{30}+\frac {8}{189} \log (2+3 x)-\frac {1}{875} \log (1+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 33, normalized size = 1.00 \begin {gather*} \frac {x^2}{30}-\frac {13 x}{225}+\frac {8}{189} \log (3 x+2)-\frac {1}{875} \log (5 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(2*x + 13*x^2 + 15*x^3),x]

[Out]

(-13*x)/225 + x^2/30 + (8*Log[2 + 3*x])/189 - Log[1 + 5*x]/875

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^4}{2 x+13 x^2+15 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^4/(2*x + 13*x^2 + 15*x^3),x]

[Out]

IntegrateAlgebraic[x^4/(2*x + 13*x^2 + 15*x^3), x]

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fricas [A]  time = 0.40, size = 25, normalized size = 0.76 \begin {gather*} \frac {1}{30} \, x^{2} - \frac {13}{225} \, x - \frac {1}{875} \, \log \left (5 \, x + 1\right ) + \frac {8}{189} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(15*x^3+13*x^2+2*x),x, algorithm="fricas")

[Out]

1/30*x^2 - 13/225*x - 1/875*log(5*x + 1) + 8/189*log(3*x + 2)

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giac [A]  time = 0.15, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{30} \, x^{2} - \frac {13}{225} \, x - \frac {1}{875} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) + \frac {8}{189} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(15*x^3+13*x^2+2*x),x, algorithm="giac")

[Out]

1/30*x^2 - 13/225*x - 1/875*log(abs(5*x + 1)) + 8/189*log(abs(3*x + 2))

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maple [A]  time = 0.07, size = 26, normalized size = 0.79 \begin {gather*} \frac {x^{2}}{30}-\frac {13 x}{225}-\frac {\ln \left (5 x +1\right )}{875}+\frac {8 \ln \left (3 x +2\right )}{189} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(15*x^3+13*x^2+2*x),x)

[Out]

1/30*x^2-13/225*x-1/875*ln(5*x+1)+8/189*ln(3*x+2)

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maxima [A]  time = 0.84, size = 25, normalized size = 0.76 \begin {gather*} \frac {1}{30} \, x^{2} - \frac {13}{225} \, x - \frac {1}{875} \, \log \left (5 \, x + 1\right ) + \frac {8}{189} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(15*x^3+13*x^2+2*x),x, algorithm="maxima")

[Out]

1/30*x^2 - 13/225*x - 1/875*log(5*x + 1) + 8/189*log(3*x + 2)

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mupad [B]  time = 0.03, size = 21, normalized size = 0.64 \begin {gather*} \frac {8\,\ln \left (x+\frac {2}{3}\right )}{189}-\frac {13\,x}{225}-\frac {\ln \left (x+\frac {1}{5}\right )}{875}+\frac {x^2}{30} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(2*x + 13*x^2 + 15*x^3),x)

[Out]

(8*log(x + 2/3))/189 - (13*x)/225 - log(x + 1/5)/875 + x^2/30

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sympy [A]  time = 0.12, size = 27, normalized size = 0.82 \begin {gather*} \frac {x^{2}}{30} - \frac {13 x}{225} - \frac {\log {\left (x + \frac {1}{5} \right )}}{875} + \frac {8 \log {\left (x + \frac {2}{3} \right )}}{189} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(15*x**3+13*x**2+2*x),x)

[Out]

x**2/30 - 13*x/225 - log(x + 1/5)/875 + 8*log(x + 2/3)/189

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